How can I prove the following inequality:
Given $ a,b>0 $ and $a^2>b $, we have $a>\sqrt b$
Thank you.
Asked
Active
Viewed 2,729 times
6
-
3Note: the answer is highly dependent on how much you already know about inequalities. – Alex Becker Mar 21 '12 at 16:25
2 Answers
21
$a^2 > b \Leftrightarrow (a - \sqrt{b})(a + \sqrt{b}) > 0$
Both of these factors must be positive, since both $a$ and $\sqrt{b}$ are positive. In particular, $a - \sqrt{b} > 0$
Indeed, I stand on the shoulders of giants...
The Chaz 2.0
- 10,464
-
Both of these factors multiplied with each other gives us positive- but why can't each of this factor be negative?(negative multiply negative gives us positive) – Anonymous Mar 21 '12 at 16:38
-
$a$ is positive. $\sqrt{b}$ is positive. When you add them, you get the positive number $ a + \sqrt{b}$, which is the second factor. So the first factor must also be positive. – The Chaz 2.0 Mar 21 '12 at 16:41
-
1
-
1
6
Suppose otherwise, i.e. that $a\leq \sqrt{b}$. Then $a^2=a\cdot a\leq \sqrt{b}\cdot a\leq \sqrt{b}\cdot\sqrt{b}=b$, so $a^2\leq b$, contradicting the fact that $a^2>\sqrt{b}$.
Alex Becker
- 60,569
-
-
BTW, is $\sqrt{b}\cdot\sqrt{b}=b$ by definition or can it be proven? – Anonymous Mar 21 '12 at 17:00
-